Balancing Equations and Reacting Masses

The law of conservation of mass states that no atoms are created or destroyed in a chemical reaction. The total mass of the reactants always equals the total mass of the products.

This means that a balanced equation must have the same number of each type of atom on both sides.

Step-by-Step Method

1. Write the correct formulae for all reactants and products
2. Count the atoms of each element on both sides
3. Add coefficients (big numbers in front) to balance each element
4. Check that all elements are balanced
5. Never change the formulae (small subscript numbers) — only add coefficients

Example 1: Hydrogen + Oxygen → Water

Unbalanced: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$

Balance oxygen by putting 2 in front of $\text{H}_2\text{O}$: $\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Now check H: Left = 2, Right = 4 ✗. Put 2 in front of $\text{H}_2$: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Check: H — 4 each side ✓; O — 2 each side ✓ ✓

Example 2: Iron + Oxygen → Iron(III) Oxide

$\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$

Balance Fe: $2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$

Balance O (need 3 on left, have 2 — use 3/2, then multiply through): $4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$

Check: Fe — 4 each side ✓; O — 6 each side ✓

Example 3: Aluminium + Hydrochloric Acid

$\text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$

Balance Cl: $\text{Al} + 3\text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$

Check H: Left = 3, Right = 2 ✗. Multiply everything: $2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2$

Check: Al — 2 ✓; H — 6 ✓; Cl — 6 ✓

Once an equation is balanced, you can use it to calculate masses.

The Method

1. Write the balanced equation
2. Calculate the $M_r$ of the substances you're interested in
3. Use the equation to find the molar ratio
4. Calculate moles of the substance you know: $n = \frac{m}{M_r}$
5. Use the ratio to find moles of the substance you want
6. Calculate the mass: $m = n \times M_r$

In most reactions, one reactant is used up completely (the limiting reactant) while the other is in excess (some is left over).

The limiting reactant determines how much product is formed.

How to Identify the Limiting Reactant

1. Calculate moles of each reactant
2. Compare the ratio of moles to the ratio in the equation
3. The reactant that runs out first is the limiting reactant

Example

Question: 4.8 g of Mg reacts with 4.0 g of O₂. Which is the limiting reactant?

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$

Moles: $n(\text{Mg}) = \frac{4.8}{24} = 0.2$ mol; $n(\text{O}_2) = \frac{4.0}{32} = 0.125$ mol

From equation: 2 mol Mg needs 1 mol O₂, so 0.2 mol Mg needs 0.1 mol O₂.

We have 0.125 mol O₂ (more than 0.1), so O₂ is in excess. Mg is the limiting reactant.

In practice, reactions rarely produce the theoretical maximum amount of product. The percentage yield measures how successful a reaction was:

$\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Reasons for Less Than 100% Yield

• The reaction is reversible (doesn't go to completion)
• Some product is lost during purification (filtering, transferring)
• Side reactions produce unwanted products
• The reaction may be incomplete (not enough time/energy)

Example

Question: The theoretical yield of CaO is 28 g, but only 22.4 g was produced. Calculate the percentage yield.

$\text{Percentage yield} = \frac{22.4}{28} \times 100 = 80\%$

Atom economy measures how much of the reactant atoms end up in the desired product:

$\text{Atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$

High atom economy is better for:

• Sustainability — less waste
• Cost — fewer wasted raw materials
• Environment — fewer by-products to dispose of

Example

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

Desired product: CaO ($M_r = 56$) All products: CaO ($56$) + CO₂ ($44$) = $100$

$\text{Atom economy} = \frac{56}{100} \times 100 = 56\%$

• Show every step in calculations — even simple ones earn marks
• Always start with a balanced equation
• Write the molar ratio clearly above the equation
• Check your answer makes sense — the product mass can't be more than the total reactant mass
• For limiting reactant questions, calculate moles of both reactants first

• Balanced equations have equal atoms of each element on both sides
• Reacting mass method: balanced equation → moles → mass
• $n = \frac{m}{M_r}$
• Limiting reactant = the reactant that runs out first
• Percentage yield = $\frac{\text{actual}}{\text{theoretical}} \times 100$
• Atom economy = $\frac{M_r \text{ of desired product}}{\text{sum of all product } M_r} \times 100$